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\(\sqrt{\dfrac{a+b}{c+ab}}+\sqrt{\dfrac{b+c}{a+bc}}+\sqrt{\dfrac{c+a}{b+ac}}\)
Bài này có xuất hiện rồi ,you vào mục tìm kiếm là thấy liền.
Lời giải vắn tắt:
\(A=\sum\sqrt{\dfrac{ab+2c^2}{1+ab-c^2}}=\sum\dfrac{ab+2c^2}{\sqrt{\left(ab+2c^2\right)\left(1+ab-c^2\right)}}\ge\sum\dfrac{2\left(ab+2c^2\right)}{1+2ab+c^2}=\sum\dfrac{2\left(ab+2c^2\right)}{\left(a+b\right)^2+2c^2}\ge\sum\dfrac{2\left(ab+2c^2\right)}{2\left(a^2+b^2\right)+2c^2}=\sum\left(ab+2c^2\right)=ab+bc+ca+2\)
( thay \(a^2+b^2+c^2=1\))
Ta có:
1+a2 = ab+bc+ca+a2 = a(a+b)+c(a+b)=(a+b)(a+c)
Tương tự: 1+b2 = (b+c)(b+a)
1+c2 = (c+a)(c+b)
\(\Rightarrow\) P = \(2a\sqrt{\dfrac{1}{\left(a+b\right)\left(a+c\right)}}+2b\sqrt{\dfrac{1}{\left(b+c\right)\left(b+a\right)}}+2c\sqrt{\dfrac{1}{\left(c+a\right)\left(c+b\right)}}\)
Áp dụng BĐT Cô-si ta có:
P\(\le\)\(a\left(\dfrac{1}{a+b}+\dfrac{1}{a+c}\right)+b\left(\dfrac{1}{4\left(b+c\right)}+\dfrac{1}{b+a}\right)+c\left(\dfrac{1}{4\left(c+b\right)}+\dfrac{1}{c+a}\right)\)\(\le\)\(\dfrac{a}{a+b}+\dfrac{a}{a+c}+\dfrac{b}{4\left(b+c\right)}+\dfrac{b}{b+a}+\dfrac{c}{4\left(c+b\right)}+\dfrac{c}{c+a}\)
= \(\dfrac{1}{4}+2=\dfrac{9}{4}\)
\(\Rightarrow\)Pmin = \(\dfrac{9}{4}\)
Dấu "=" xảy ra\(\Leftrightarrow\) b=c=\(\dfrac{a}{7}\)=\(\dfrac{\sqrt{15}}{15}\) \(\Rightarrow\) a = \(\dfrac{7\sqrt{15}}{15}\)
a)Bunhia:
\(\left(1+2\right)\left(b^2+2a^2\right)\ge\left(1.b+\sqrt{2}.\sqrt{2}a\right)^2=\left(b+2a\right)^2\)
b)\(ab+bc+ca=abc\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\)
Áp dụng bđt câu a
=>VT\(\ge\)\(\dfrac{b+2a}{\sqrt{3}ab}+\dfrac{c+2b}{\sqrt{3}bc}+\dfrac{a+2c}{\sqrt{3}ca}\)
\(\Leftrightarrow VT\ge\dfrac{1}{a}+\dfrac{2}{b}+\dfrac{1}{b}+\dfrac{2}{c}+\dfrac{1}{c}+\dfrac{2}{a}=3=VP\)
Tự tìm dấu "="
Nguyễn Việt LâmMashiro ShiinaBNguyễn Thanh HằngonkingCẩm MịcFa CTRẦN MINH HOÀNGhâu DehQuân Tạ MinhTrương Thị Hải Anh
Ta có: $\sqrt[]{ab+2c}=\sqrt[]{ab+(a+b+c)c}=\sqrt[]{ab+ac+bc+c^2}=\sqrt[]{(c+a)(c+b)}$ (do $a+b+c=2$)
Nên $\dfrac{ab}{\sqrt[]{ab+2c}}=\dfrac{ab}{\sqrt[]{(c+a).(c+b)}}=ab.\sqrt[]{\dfrac{1}{a+c}.\dfrac{1}{b+c}}$
Áp dụng bất đẳng thức Cauchy cho $\dfrac{1}{a+c};\dfrac{1}{b+c}>0$ có:
$\sqrt[]{\dfrac{1}{a+c}.\dfrac{1}{b+c}} \leq \dfrac{1}{2}.(\dfrac{1}{a+c}+\dfrac{1}{b+c})$
Nên $\dfrac{ab}{\sqrt[]{ab+2c}} \leq \dfrac{1}{2}.ab.(\dfrac{1}{a+c}+\dfrac{1}{b+c})= \dfrac{1}{2}.(\dfrac{ab}{a+c}+\dfrac{ab}{b+c})$
Tương tự ta có: $\dfrac{bc}{\sqrt[]{bc+2a}} \leq \dfrac{1}{2}.(\dfrac{bc}{a+b}+\dfrac{bc}{a+c})$
$\dfrac{ca}{\sqrt[]{ca+2b}} \leq \dfrac{1}{2}.(\dfrac{ca}{b+a}+\dfrac{ca}{b+c})$
Nên $Q \leq \dfrac{1}{2}.(\dfrac{ab}{a+c}+\dfrac{ab}{b+c})+\dfrac{1}{2}.(\dfrac{bc}{a+b}+\dfrac{bc}{a+c})+ \dfrac{1}{2}.(\dfrac{ca}{b+a}+\dfrac{ca}{b+c})=\dfrac{1}{2}(\dfrac{ab}{a+c}+\dfrac{ab}{b+c}+\dfrac{bc}{a+b}+\dfrac{bc}{a+c}+\dfrac{ca}{b+a}+\dfrac{ca}{b+c})=\dfrac{1}{2}.(\dfrac{b(a+c)}{a+c}+\dfrac{a(b+c)}{b+c}+\dfrac{c(a+b)}{a+b}=\dfrac{1}{2}.(a+b+c)=1$ (do $a+b+c=2$)
Dấu $=$ xảy ra khi $a=b=c=\dfrac{2}{3}$
Lợi dụng Cauchy-Schwarz' inequality ta có:
\(\dfrac{ab}{\sqrt{ab+2c}}=\dfrac{ab}{\sqrt{ab+\left(a+b+c\right)c}}=\dfrac{ab}{\sqrt{ab+ac+bc+c^2}}\)
\(=\dfrac{ab}{\sqrt{\left(a+c\right)\left(b+c\right)}}\le\dfrac{1}{2}\left(\dfrac{ab}{a+c}+\dfrac{ab}{b+c}\right)\)
Tương tự ta cũng có:
\(\dfrac{bc}{\sqrt{bc+2a}}\le\dfrac{1}{2}\left(\dfrac{bc}{a+b}+\dfrac{bc}{a+c}\right);\dfrac{ca}{\sqrt{ca+2b}}\le\dfrac{1}{2}\left(\dfrac{ca}{a+b}+\dfrac{ca}{b+c}\right)\)
Cộng theo vế 3 BĐT trên ta có:
\(P\le\dfrac{1}{2}\left(\dfrac{ab+bc}{a+c}+\dfrac{bc+ca}{a+b}+\dfrac{ab+ca}{b+c}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{b\left(a+c\right)}{a+c}+\dfrac{c\left(a+b\right)}{a+b}+\dfrac{a\left(b+c\right)}{b+c}\right)\)
\(=\dfrac{1}{2}\left(a+b+c\right)=\dfrac{1}{2}\cdot2=1\)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{2}{3}\)
Ta có P=\(\dfrac{ab}{\sqrt{ab+\left(a+b+c\right)c}}+\dfrac{bc}{\sqrt{bc+\left(a+b+c\right)a}}+\dfrac{ac}{\sqrt{ac+\left(a+b+c\right)b}}\)
=\(\dfrac{ab}{\sqrt{ab+ac+bc+c^2}}+\dfrac{bc}{\sqrt{bc+ac+ab+a^2}}+\dfrac{ac}{\sqrt{ac+ab+bc+b^2}}\)
=\(\dfrac{ab}{\sqrt{a\left(b+c\right)+c\left(b+c\right)}}+\dfrac{bc}{\sqrt{b\left(a+c\right)+a\left(a+c\right)}}+\dfrac{ac}{\sqrt{c\left(a+b\right)+b\left(a+b\right)}}\)
=\(\dfrac{ab}{\sqrt{\left(a+c\right)\left(b+c\right)}}+\dfrac{bc}{\sqrt{\left(b+a\right)\left(c+a\right)}}+\dfrac{ac}{\sqrt{\left(a+b\right)\left(c+b\right)}}\)
áp dụng bđt Cói ta có:
\(\sqrt{\left(a+c\right)\left(b+c\right)}\)\(\le\)\(\dfrac{2+c}{2}=1+\dfrac{c}{2}\)
\(\sqrt{\left(b+á\right)\left(c+a\right)}\)
Áp dụng BĐT phụ:
\(3\left(a^2+a^2+b^2\right)\ge\left(2a+b\right)^2\)
P=\(\sum\dfrac{a}{\sqrt{2a^2+b^2}+\sqrt{3}}\)
\(\Rightarrow\)\(\dfrac{1}{\sqrt{3}}P=\sum\dfrac{a}{\sqrt{3\left(a^2+a^2+b^2\right)}+3}\)
\(\Rightarrow\)\(\dfrac{1}{\sqrt{3}}P\le\sum\dfrac{a}{\sqrt{\left(2a+b\right)^2}+a+b+c}=\sum\dfrac{a}{3a+2b+c}\)
Xét M=\(\sum\dfrac{a}{3a+2b+c}\)
\(3-3M=\sum\dfrac{2b+c}{3a+2b+c}\)
\(\Rightarrow\)\(3-3M=\sum\dfrac{\left(2b+c\right)^2}{\left(3a+2b+c\right)\left(2b+c\right)}\ge\)\(\dfrac{\left(3a+3b+3c\right)^2}{\sum\left(3a+2b+c\right)\left(2b+c\right)}\)
Mà
\(\sum\left(3a+2b+c\right)\left(2b+c\right)=5a^2+5b^2+5c^2+13ab+13bc+13ac=5\left(a+b+c\right)^2+3\left(ab+bc+ac\right)\le5\left(a+b+c\right)^2+\left(a+b+c\right)^2\)
\(\Rightarrow\)\(3-3M\ge\dfrac{\left(3a+3b+3c\right)^2}{6\left(a+b+c\right)^2}\ge\dfrac{9}{6}=\dfrac{3}{2}\)
\(\Rightarrow\)\(M\le\dfrac{1}{2}\)
\(\Rightarrow\)\(\dfrac{1}{\sqrt{3}}P\le\dfrac{1}{2}\Rightarrow P\le\dfrac{\sqrt{3}}{2}\)
Bài 2:
\(\sqrt{\dfrac{a}{b+c}}+\sqrt{\dfrac{b}{c+a}}+\sqrt{\dfrac{c}{a+b}}>2\)
Trước hết ta chứng minh \(\sqrt{\dfrac{a}{b+c}}\ge\dfrac{2a}{a+b+c}\)
Áp dụng BĐT AM-GM ta có:
\(\sqrt{a\left(b+c\right)}\le\dfrac{a+b+c}{2}\)\(\Rightarrow1\ge\dfrac{2\sqrt{a\left(b+c\right)}}{a+b+c}\)
\(\Rightarrow\sqrt{\dfrac{a}{b+c}}\ge\dfrac{2a}{a+b+c}\). Ta lại có:
\(\sqrt{\dfrac{a}{b+c}}=\dfrac{\sqrt{a}}{\sqrt{b+c}}=\dfrac{a}{\sqrt{a\left(b+c\right)}}\ge\dfrac{2a}{a+b+c}\)
Thiết lập các BĐT tương tự:
\(\sqrt{\dfrac{b}{c+a}}\ge\dfrac{2b}{a+b+c};\sqrt{\dfrac{c}{a+b}}\ge\dfrac{2c}{a+b+c}\)
Cộng theo vế 3 BĐT trên ta có:
\(VT\ge\dfrac{2a}{a+b+c}+\dfrac{2b}{a+b+c}+\dfrac{2c}{a+b+c}=\dfrac{2\left(a+b+c\right)}{a+b+c}\ge2\)
Dấu "=" không xảy ra nên ta có ĐPCM
Lưu ý: lần sau đăng từng bài 1 thôi nhé !
1) Áp dụng liên tiếp bđt \(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\) với a;b là 2 số dương ta có:
\(\dfrac{1}{2a+b+c}=\dfrac{1}{\left(a+b\right)+\left(a+c\right)}\le\dfrac{\dfrac{1}{a+b}+\dfrac{1}{a+c}}{4}\)\(\le\dfrac{\dfrac{2}{a}+\dfrac{1}{b}+\dfrac{1}{c}}{16}\)
TT: \(\dfrac{1}{a+2b+c}\le\dfrac{\dfrac{2}{b}+\dfrac{1}{a}+\dfrac{1}{c}}{16}\)
\(\dfrac{1}{a+b+2c}\le\dfrac{\dfrac{2}{c}+\dfrac{1}{a}+\dfrac{1}{b}}{16}\)
Cộng vế với vế ta được:
\(\dfrac{1}{2a+b+c}+\dfrac{1}{a+2b+c}+\dfrac{1}{a+b+2c}\le\dfrac{1}{16}.\left(\dfrac{4}{a}+\dfrac{4}{b}+\dfrac{4}{c}\right)=1\left(đpcm\right)\)
bai nay t lam roi vao trang chu cua nick thangbnsh cua t keo xuong tim la thay
Câu hỏi của Tuyển Trần Thị - Toán lớp 9 | Học trực tuyến
Câu này giải như sau :
Ta có :
\(\sqrt{2a+bc}=\sqrt{\left(a+b+c\right)a+bc}=\sqrt{a^2+ab+ac+bc}=\sqrt{\left(a+c\right)\left(a+b\right)}\)
\(\Rightarrow\sqrt{2a+bc}\le\frac{a+b+a+c}{2}=\frac{2a+b+c}{2}\left(1\right)\)
tương tự ta có :\(\sqrt{2b+ac}\le\frac{2b+a+c}{2}\left(2\right)\)
\(\sqrt{2c+ac}\le\frac{2c+a+c}{2}\left(3\right)\)
cộng vế với vế 1,2,3 ta được
\(Q\le\frac{3\left(a+b+c\right)}{2}=\frac{3.2}{2}=3\)\(\Rightarrow Q_{max}=3\Leftrightarrow\)dấu "=" (a,b,c) là hoán vị của \(\left(0.1.1\right)\)
Lời giải:
Áp dụng BĐT AM-GM:
\(P=\frac{\sqrt{ab}}{(a+c)+(b+c)}+\frac{\sqrt{bc}}{(b+a)+(c+a)}+\frac{\sqrt{ca}}{(c+b)+(a+b)}\)
\(\leq \underbrace{\frac{\sqrt{ab}}{2\sqrt{(a+c)(b+c)}}+\frac{\sqrt{bc}}{2\sqrt{(b+a)(c+a)}}+\frac{\sqrt{ca}}{2\sqrt{(c+b)(a+b)}}}_{M}(*)\)
Xét:
\(M=\frac{1}{2}\frac{\sqrt{ab(a+b)}+\sqrt{bc(b+c)}+\sqrt{ca(c+a)}}{\sqrt{(a+b)(b+c)(c+a)}}(1)\)
Theo BĐT Bunhiacopxky và AM-GM:
\((\sqrt{ab(a+b)}+\sqrt{bc(b+c)}+\sqrt{ca(c+a)})^2\leq (ab+bc+ac)(a+b+b+c+c+a)\)
\(=2(ab+bc+ac)(a+b+c)=2[(a+b)(b+c)(c+a)+abc]\)
\(\leq 2[(a+b)(b+c)(c+a)+\frac{(a+b)(b+c)(c+a)}{8}]=\frac{9}{4}(a+b)(b+c)(c+a)\)
\(\Rightarrow \sqrt{ab(a+b)}+\sqrt{bc(b+c)}+\sqrt{ca(c+a)}\leq \frac{3}{2}\sqrt{(a+b)(b+c)(c+a)}(2)\)
Từ \((1);(2)\Rightarrow M\leq \frac{1}{2}.\frac{3}{2}=\frac{3}{4}(**)\)
Từ \((*); (**)\Rightarrow P\leq M\leq \frac{3}{4}\)
Vậy \(P_{\max}=\frac{3}{4}\Leftrightarrow a=b=c\)
em cảm ơn cô